By Steel5
Date 11-16-2017 22:43
Edited 11-16-2017 23:31
Ran across this thread looking for information on flare-bevel groove welds, this is a great discussion! Sorry for posting on a somewhat older topic, hope you guys can provide me with some insights.
I’d like to ask a question regarding Al’s last paragraph starting with “the attached sketch…”.
I think I understand the general concept: flare bevel weld should be flush, but if it’s less than flush you can discount some of the weld and still calculate some strength out of it.
Where my question lies is in what the “Total Weld Size (E)” would be at the end. I follow your calculation and sketch, but it’s written in terms of “effective weld size”, which I don’t think is the same as “Total Weld Size (E)”.
I typically work out of the AISC Manual 14th edition and have less familiarity with AWS D1.1.
I found an old edition of AWS D1.1 from 1996, and found table 3.5 (which is actually 3.4 in this version) and table 2.1, so I see where you’re grabbing the values from, but I’m concerned with the “Weld Size (E)” listed in the prequalified weld tables, which in this 1996 version is figure 3.3 and Flare-bevel-groove weld is on pg.65.
These prequalified weld tables are reproduced in AISC Manual table 8-2, which is where I most commonly refer to them. In both the AISC manual (pg.8-61) and AWS D1.1 (pg.65 in old edition), the “Weld Size (E)” for a flare-bevel groove weld is 5*T1/8. Interestingly, it’s not even a function of radius. For the example you gave of a 1in thick HSS, the “Weld Size (E)” would be 5*1in/8 = 5/8in.
One thing I can initially conclude is that if the weld is flush, I have the full value of (E)=5/8in.
But what if the weld is not flush? Say it’s just like your example, where the “effective weld size” was supposed to be 0.47in but is now actually 0.31in.
One method might be to look at the difference, which is 0.16in, and subtract that from (E) to get 5/8in – 0.16in = 0.465in. But this seems erroneous because then if the “effective weld” were reduce to 0in, then (E) would still be 5/8in-0.47in=0.155in by that logic, which doesn’t make sense.
A better way to look at it might be to use a ratio. 0.31in/0.47in = 0.66 = 66%. So perhaps the “Weld size (E)” value would be (E)=5/8in * 0.66 = 0.413in. Seems to make more intuitive sense, but it’s just an idea, I’d like to know what you guys think.
Generally, I’m picturing this whole problem from an evaluating an existing condition perspective, but I also wonder what this would be like from a design perspective. If I specify a (E)=(1/2in), how does the welder determine how much to weld?
Hope my question makes sense. Ultimately the value that’s used in engineering design is the “Total Weld Size (E)”, so it’s very important to calculate correctly and understand. Thanks for any help!